
\prob{00A4}{Taylor展开}

求证：对于某实数$x_0$，可导函数$f(x)$可以表示为如下形式：
\[ f(x) = \sum_{n = 0}^\infty \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n \]
其中$f^{(n)}$表示$f$的$n$阶导数。
\problabels{yellow/微积分, green/证明题}

\subsection{待定系数}

设$f(x) = a_0(x - x_0)^0 + a_1(x - x_0)^1 + \dots + a_n(x - x_0)^n + \dots$，其中$x_0$表示在$x_0$处展开。易知$f(x_0) = a_0$。

运用链式法则将$f(x)$求导得$f'(x)$，其中
\begin{align*}
  f'(x) &= 0 + a_1\cdot1(x - x_0)^0\cdot(1 - 0) \\
  &+ a_2\cdot2(x - x_0)^1\cdot(1 - 0) \\
  &+ a_3\cdot3(x - x_0)^2\cdot(1 - 0) + \cdots \\
  &+ a_n\cdot n(x - x_0)^{n - 1}\cdot(1 - 0) + \cdots \\
  &= a_1 + 2a_2(x - x_0)^1 + 3a_3(x - x_0)^2 \\
  &+ \dots + na_n(x - x_0)^{n - 1} + \cdots
\end{align*}
由此可知$f'(x_0) = a_1$。再次求导可得
\begin{align*}
  f''(x) &= 0 + 2a_2\cdot1(x - x_0)^0\cdot(1 - 0) \\
  &+ 3a_3\cdot2(x - x_0)^1\cdot(1 - 0) + \cdots \\
  &+ na_n\cdot(n - 1)(x - x_0)^{n - 2}\cdot(1 - 0) + \cdots \\
  &= 1\cdot2a_2 + 2\cdot3a_3(x - x_0)^1 \\
  &+ 3\cdot4a_4(x - x_0)^2 + \cdots \\
  &+ (n - 1)na_n(x - x_0)^{n - 2} + \cdots
\end{align*}
易知$(1/2!)f''(x_0) = (1/(1\cdot2))f''(x_0) = a_2$。多次求导可得
\begin{align*}
  f(x) &= 0!a_0(x - x_0)^0 + \frac{1!}{1!}a_1(x - x_0)^1 \\
  &+ \dots + \frac{n!}{n!}a_n(x - x_0)^n + \cdots \\
  f'(x) &= 1!a_1(x - x_0)^0 + \frac{2!}{1!}a_2(x - x_0)^1 \\
  &+ \dots + \frac{n!}{(n - 1)!}a_n(x - x_0)^{n - 1} + \cdots \\
  f''(x) &= 2!a_2(x - x_0)^0 + \frac{3!}{1!}a_3(x - x_0)^1 \\
  &+ \dots + \frac{n!}{(n - 2)!}a_n(x - x_0)^{n - 2} + \cdots \\
  f'''(x) &= 3!a_3(x - x_0)^0 + \frac{4!}{1!}a_4(x - x_0)^1 + \cdots \\
  &+ \frac{n!}{(n - 3)!}a_n(x - x_0)^{n - 3} + \cdots \\
  f^{(k)}(x) &= k!a_k(x - x_0)^0 + \frac{(k + 1)!}{1!}a_{k + 1}(x - x_0)^1 \\
  &+ \dots + \frac{n!}{(n - k)!}a_n(x - x_0)^{n - k} + \cdots
\end{align*}
易知
\begin{align*}
  a_0 &= \frac{f(x_0)}{0!} \\
  a_1 &= \frac{f'(x_0)}{1!} \\
  \cdots &= \cdots \\
  a_n &= \frac{f^{(n)}(x_0)}{n!}
\end{align*}
由此可得$f(x)$的Taylor展开式为\footnote{当函数在$0$处展开，即$x_0 = 0$时，该展开被称为Maclaurin展开，即$f(x) = \sum_{n = 0}^\infty (f^{(n)}(0)/n!)x^n$。}
\begin{align*}
  f(x) &= \frac{f(x_0)}{0!}(x - x_0)^0 + \frac{f'(x_0)}{1!}(x - x_0)^1 \\
  &+ \dots + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n + \cdots \\
  &= \sum_{n = 0}^\infty \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n \\
\end{align*}
